Question

Tìm ĐKXĐ:

A = \(\dfrac{1}{\sqrt{x-3}}\)

B = \(\sqrt{x-2}+\dfrac{1}{\sqrt{x-2}}\)

Answer 1

a. ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{x-3}\ne0\\x-3\ge0\end{matrix}\right.\)⇌\(\left\{{}\begin{matrix}x\ne3\\x\ge3\end{matrix}\right.\)⇒ x > 3

b.ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)⇌\(\left\{{}\begin{matrix}x\ge2\\x\ne2\end{matrix}\right.\)⇒x > 2

Answer 2

a) Đk: \(\left\{{}\begin{matrix}x-3\ge0\\x-3\ne0\end{matrix}\right.\Leftrightarrow x-3>0\Leftrightarrow x>3\)

b) \(x-2>0\Leftrightarrow x>2\)

Answer 3

A=\(\dfrac{1}{\sqrt{x-3}}\)

Để A xác định thì x - 3 > 0

\(\Leftrightarrow\) x > 3